JEE Main 2025 April 8 Shift 2 Chemistry Question Paper, Exam Analysis, and Answer Keys (Available)

Analysis of Statement I:

- Acidic character increases down the group for hydrides (H\(_2\)O < H\(_2\)S \(<\) H\(_2\)Se \(<\) H\(_2\)Te)
- \( H_2Te \) is more acidic than \( H_2Se \) because:

Te is larger than Se, making Te-H bond weaker
H\(^+\) ion is more easily released from \( H_2Te \)

- Therefore, Statement I is false

Analysis of Statement II:
- Bond enthalpy decreases down the group (H-Se > H-Te)
- \( H_2Se \) has higher bond dissociation enthalpy because:

Se is smaller than Te, forming stronger bonds
Bond strength decreases with increasing atomic size

- Therefore, Statement II is true Quick Tip: - Remember the periodic trend: acidity of hydrides increases down the group - Bond enthalpy decreases with increasing atomic size in a group - Larger atoms form weaker bonds, making their hydrides more acidic

Magnetic moment (\( \mu_{spin} \)) depends on the number of unpaired electrons in the ions. The formula for the magnetic moment due to spin only is:
\[ \mu_{spin} = \sqrt{n(n+2)} \, BM \]

Where \( n \) is the number of unpaired electrons.


Step 1: Determine the number of unpaired electrons for each ion:


- \( Cu^+ \) (\( Cu^{1+} \)): The electron configuration of Cu is \( [Ar] 3d^{10} 4s^1 \). For \( Cu^+ \), the electron configuration is \( [Ar] 3d^{10} \), so there are no unpaired electrons. Thus, \( \mu_{spin} = 0 \, BM \).


- \( Cu^{2+} \): The electron configuration of \( Cu^{2+} \) is \( [Ar] 3d^9 \), which has 1 unpaired electron. Thus, \( \mu_{spin} = \sqrt{1(1+2)} = \sqrt{3} \, BM \).


- \( Cr^{2+} \): The electron configuration of \( Cr^{2+} \) is \( [Ar] 3d^4 \), which has 4 unpaired electrons. Thus, \( \mu_{spin} = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6} \, BM \).

- \( Cr^{3+} \): The electron configuration of \( Cr^{3+} \) is \( [Ar] 3d^3 \), which has 3 unpaired electrons. Thus, \( \mu_{spin} = \sqrt{3(3+2)} = \sqrt{15} \, BM \).

Step 2: Compare the magnetic moment values.


- \( Cu^+ \) has 0 unpaired electrons, so its magnetic moment is 0 BM.

- \( Cu^{2+} \) has 1 unpaired electron, so its magnetic moment is \( \sqrt{3} \, BM \).

- \( Cr^{3+} \) has 3 unpaired electrons, so its magnetic moment is \( \sqrt{15} \, BM \).

- \( Cr^{2+} \) has 4 unpaired electrons, so its magnetic moment is \( 2\sqrt{6} \, BM \).


Step 3: Final decreasing order.


Thus, the correct decreasing order of spin only magnetic moment values is:
\[ \boxed{Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^+} \] Quick Tip: The magnetic moment of transition metal ions depends on the number of unpaired electrons. A higher number of unpaired electrons results in a higher magnetic moment.

Matching Analysis:

A. Sodium bicarbonate solution - Reacts with carboxylic acids (II) to produce effervescence (CO\(_2\))
B. Neutral ferric chloride - Gives colored complex with phenolic -OH (III)
C. Ceric ammonium nitrate - Red color with alcoholic -OH (IV)
D. Alkaline KMnO\(_4\) - Decolorizes with double bonds (I)


Correct Mapping:

A → II (Carboxylic acid)
B → III (Phenolic -OH)
C → IV (Alcoholic -OH)
D → I (Double bond) Quick Tip: - Sodium bicarbonate tests for carboxylic acids via CO\(_2\) evolution - Ferric chloride gives violet color with phenols - Ceric ammonium nitrate turns red with alcohols - Baeyer's reagent (alk. KMnO\(_4\)) tests unsaturation

Analysis of Statement I:

- A homoleptic octahedral complex with monodentate ligands (e.g., \([Co(NH_3)_6]^{3+})\) has identical ligands in all positions
- No geometrical or optical isomers possible as all positions are equivalent
- Therefore, Statement I is true


Analysis of Statement II:

- cis-platin and trans-platin are actually Pt(II) complexes, not Pd
- Both are homoleptic complexes (contain same ligands: \([PtCl_2(NH_3)_2])\)
- The difference is in ligand arrangement (cis/trans), not heteroleptic nature
- Therefore, Statement II is false Quick Tip: - Homoleptic complexes have only one type of ligand - Geometrical isomerism requires different spatial arrangements - Cis/trans-platin are classic examples of geometrical (not optical) isomerism - Platinum (Pt) complexes \(\neq\) Palladium (Pd) complexes

The structure of the compound shows a cyclopentene ring with a hydroxyl group (-OH) attached to one of its carbon atoms and an ethyl group attached at position (4)


Step 1: Identify the parent chain and the functional group.

- The parent structure is a cyclopentene ring, which is a five-membered ring with one double bond.

- The double bond is at position 2, as the numbering starts from the carbon atom of the double bond.

- The hydroxyl group (-OH) is attached at position 1 of the ring.


Step 2: Name the substituents.

- An ethyl group (-C2H5) is attached at position 4 of the ring.

- The hydroxyl group is at position 1, so it is named as "1-ol."


Step 3: Final IUPAC name.

The correct IUPAC name of the compound is:

4-Ethylcyclopent-2-en-1-ol Quick Tip: When naming cyclic compounds with multiple substituents, number the ring starting from the position of the highest-priority functional group and then assign the lowest possible number to other substituents based on alphabetical order.

Let’s analyze the given reactions and clues for each compound.


Step 1: Analyzing A:

- A shows positive Lassaigne’s test for nitrogen (N), which means A contains nitrogen.

- The molar mass of A is 121 g/mol.

- Based on the structure options, the compound with a molecular formula that fits these criteria is \(CONH_2\) (amide group). This is the structure of \(benzamide\), which fits the molar mass and the positive Lassaigne’s test for nitrogen.


Step 2: Analyzing B:

- B gives effervescence with aqueous NaHCO(3)This suggests B contains a carboxyl group (-COOH), as carboxylic acids react with NaHCO3 to release carbon dioxide (CO2).

- Therefore, B corresponds to a carboxylic acid group, specifically \(CO_2H\), which matches the second compound in the options.

Step 3: Analyzing C:

- C gives a fruity smell, which is characteristic of an ester group.

- The ester functional group in the options is \(CO_2Et\) (ethyl ester), which corresponds to the third structure in the options.

Step 4: Final Identification:

- A is \(CONH_2\) (benzamide),
- B is \(CO_2H\) (carboxylic acid),
- C is \(CO_2Et\) (ethyl ester).

Thus, the correct answer is Option (1):
A = CONH2, B = CO2H, C = CO2Et Quick Tip: When solving organic chemistry questions, pay close attention to functional groups and their specific reactions, such as effervescence with NaHCO3 (carboxylic acids) and fruity smells (esters).

Step 1: Calculate mass of hydrogen in \( H_2O \) \[ Mass of H = \frac{2}{18} \times 0.127\, g = 0.0141\, g \] \[ % H = \left( \frac{0.0141}{0.210} \right) \times 100 = 6.72% \]

Step 2: Calculate mass of carbon in \( CO_2 \) \[ Mass of C = \frac{12}{44} \times 0.307\, g = 0.0837\, g \] \[ % C = \left( \frac{0.0837}{0.210} \right) \times 100 = 39.87% \]

Step 3: Calculate percentage of oxygen \[ % O = 100 - (% C + % H) = 100 - (39.87 + 6.72) = 53.41% \] Quick Tip: - For combustion analysis: All H converts to \( H_2O \) (2 g H per 18 g \( H_2O \)) All C converts to \( CO_2 \) (12 g C per 44 g \( CO_2 \)) Oxygen % is obtained by difference - Always verify that percentages sum to 100%

Step 1: Calculate van't Hoff factor (i) \[ \Delta T_f = i \cdot K_f \cdot m
0.20 = i \times 1.8 \times 0.1
i = \frac{0.20}{0.18} = 1.11 \]

Step 2: Relate i to degree of dissociation \((\alpha)\)
For weak acid dissociation: \[ i = 1 + \alpha
1.11 = 1 + \alpha
\alpha = 0.11 \]

Step 3: Calculate dissociation constant \((K_a)\) \[ K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.1 \times (0.11)^2}{1 - 0.11}
= \frac{0.1 \times 0.0121}{0.89} = 1.36 \times 10^{-3} \approx 1.38 \times 10^{-3} \] Quick Tip: - For weak electrolytes: \( i = 1 + (n-1)\alpha \) (n = ions produced) - Freezing point depression: \( \Delta T_f = iK_fm \) - \( K_a \) calculation for weak acid: \( K_a = \frac{C\alpha^2}{1-\alpha} \) - Approximation valid when \( \alpha < 0.1 \), otherwise use exact formula

Correct Matching:

A. Carbocation - III (sp\(^2\) hybridized carbon with empty p-orbital)
B. C-Free radical - IV (sp\(^2\)/sp\(^3\) hybridized carbon with one unpaired electron)
C. Nucleophile - I (Species that can supply a pair of electrons)
D. Electrophile - II (Species that can receive a pair of electrons)


Explanation:

Carbocations are electron-deficient species with sp\(^2\) hybridization and empty p-orbital
Free radicals have one unpaired electron and can be sp\(^2\) or sp\(^3\) hybridized
Nucleophiles are electron-rich species that donate electron pairs
Electrophiles are electron-deficient species that accept electron pairs Quick Tip: - Carbocation: \(sp^2\), 6 electrons, planar - Free radical: 7 electrons, can be \(sp^2\) or \(sp^3\) - Nucleophile: "nucleus loving" (e\(^-\) donor) - Electrophile: "electron loving" (e\(^-\) acceptor)

The reactions proceed as follows:


Step 1: In the first step, \(KOH (alc.)\) induces an elimination reaction, which removes a bromine atom from the carbon chain, resulting in the formation of cyclooctene (a 8-membered ring with a double bond).


Step 2: The second step with \(NaNH_2\) induces further elimination, removing another hydrogen atom from the carbon-carbon single bonds (a step leading to the formation of cyclooctyne, a 8-membered ring with a triple bond).


Step 3: The reaction with \(Hg^{2+}/H^{+}\) induces hydromercuration of the alkyne, resulting in the addition of a mercury ion to the carbon-carbon triple bond. This creates an alkene intermediate.


Step 4: Finally, the Clemmensen reduction with \(Zn-Hg / H^{+}\) reduces the alkene to a fully saturated ring, cyclooctane, which is the final product. This reaction removes the double bond and saturates the 8-membered ring.
Quick Tip: When dealing with elimination and reduction reactions, it's essential to understand how the reagents affect the intermediates and lead to the final product. Alcohol formation and reduction are key steps in this type of organic transformation.

Step 1: Recall first-order kinetics equation \[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]
where \(k\) is the rate constant, \([A]_0\) is initial concentration, and \([A]\) is concentration at time \(t\).

Step 2: Calculate \(t_1\) for 1/4th decomposition \[ t_1 = \frac{2.303}{k} \log \frac{1}{1/4} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} \times 2 \log 2 \]

Step 3: Calculate \(t_2\) for 1/8th decomposition \[ t_2 = \frac{2.303}{k} \log \frac{1}{1/8} = \frac{2.303}{k} \log 8 = \frac{2.303}{k} \times 3 \log 2 \]

Step 4: Find the ratio \(t_1/t_2\) \[ \frac{t_1}{t_2} = \frac{2 \log 2}{3 \log 2} = \frac{2}{3} \]

However, the question asks for decomposition to one fourth (remaining = 3/4) and one eighth (remaining = 7/8). The correct interpretation is:

Correct Calculation:
For decomposition to:
- 1/4 remaining: \( t_1 = \frac{2.303}{k} \log 4 \)
- 1/8 remaining: \( t_2 = \frac{2.303}{k} \log 8 \)

Thus: \[ \frac{t_1}{t_2} = \frac{\log 4}{\log 8} = \frac{2 \log 2}{3 \log 2} = \frac{2}{3} \] Quick Tip: - For first-order reactions, time depends on the fraction remaining, not the absolute concentration - Each half-life period is equal for first-order reactions - The time to reach 1/4 is less than time to reach 1/8 (\(\frac{2}{3}\) ratio)

Correct Matching:

A. \([Ni(CO)_4]\) - III (Tetrahedral, 0 BM)

CO is strong field ligand → low spin complex
\(d^8\) configuration → tetrahedral with paired electrons \((\mu = 0)\)


B. \([Ni(CN)_4]^{2-}\) - II (Square planar, 0 BM)

\(CN^-\) is strong field ligand → low spin
\(d^8\) configuration → square planar with paired electrons s \((\mu = 0)\)


C. \([NiCl_4]^{2-}\) - I (Tetrahedral, 2.8 BM)

\(Cl^{⁻}\) is weak field ligand high spin
\(d^8\) configuration → tetrahedral with 2 unpaired electrons \((\mu = \sqrt{8} \neq 2.8 BM)\)


D. \([MnBr_4]^{2-}\) - IV (Tetrahedral, 5.9 BM)

Br⁻ is weak field ligand → high spin
d⁵ configuration → tetrahedral with 5 unpaired electrons (μ = √35 ≈ 5.9 BM) Quick Tip: - Strong field ligands (CO, CN⁻) cause pairing → low spin complexes - Weak field ligands (Cl⁻, Br⁻) don't cause pairing → high spin complexes - Magnetic moment (μ) = √[n(n+2)] BM, where n = unpaired electrons - Ni²⁺ (d⁸): tetrahedral/square planar; Mn²⁺ (d⁵): always high spin

In Option 3, the reaction involves iso-butyl alcohol (iso-BuOH) and sec-butyl bromide (sec-BuBr) in the presence of Na. This is an example of the Williamson Ether Synthesis, which proceeds through an SN2 mechanism, where a nucleophile attacks an electrophilic carbon.


However, iso-butyl (iso-Bu) is a secondary alkyl group and exhibits significant steric hindrance. This steric hindrance makes the iso-butyl group less reactive in the nucleophilic substitution reaction compared to other alkyl groups like sec-butyl. The sec-butyl group (sec-Bu) is much more reactive in this case, making it more likely to participate in the reaction and thus, the iso-butyl ether is not the major product. Therefore, Option 3 will not lead to the desired ether in major proportion. Quick Tip: When performing Williamson Ether Synthesis, consider the steric hindrance of the alkyl groups. The reaction is more favorable when the nucleophile is less hindered. Secondary and tertiary alkyl groups show less reactivity in comparison to primary alkyl groups.

Element Analysis:
- Atomic number 9 is Fluorine (\( F \))
- Electronic configuration: \( 1s^2 2s^2 2p^5 \)
- Orbital diagram:
\( 1s^2 \uparrow\downarrow \quad 2s^2 \uparrow\downarrow \quad 2p^5 \uparrow\downarrow \uparrow\downarrow \uparrow \)

Statement Evaluation:

A: TRUE
- Total electrons = 9
- 5 with \( m_s = +\frac{1}{2} \) (all unpaired + one paired)
- 4 with \( m_s = -\frac{1}{2} \) (remaining paired electrons)

B: FALSE
- \( p_z \) orbital contains 1 electron (\(\uparrow\)), but same applies to \( p_x \) and \( p_y \)
- Not a unique characteristic

C: TRUE
- Last electron enters \( 2p \) orbital (\( n=2 \), \( l=1 \))

D: FALSE
- Angular nodes = \( l \)
- Sum: \( 1s \) (0) + \( 2s \) (0) + \( 2p \) (1 each \(\times\) 3) = 3 \(\neq\) 1 Quick Tip: - For p-block elements: - \( n \) = principal quantum number - \( l \) = 1 for p-orbitals - Angular nodes = \( l \) - \( m_s \) distribution follows Hund's rule - Fluorine has 1 unpaired electron in \( 2p \) subshell

Step 1: Identify \( sp^3d^2 \) hybridization criteria
- Occurs in octahedral complexes/compounds
- Requires 6 empty orbitals (1 \( s \), 3 \( p \), 2 \( d \))
- Typical for species with either:

Central atom with 6 bond pairs (e.g., \( SF_6 \))
Transition metal complexes with weak field ligands


Step 2: Analyze each species

\( [Co(NH_3)_6]^{3+} \):
- \( NH_3 \) is strong field ligand → \( d^2sp^3 \) (inner orbital)
- Not \( sp^3d^2 \)

\( SF_6 \):
- \( S \) uses \( sp^3d^2 \) hybridization (6 bond pairs)
- Count = 1

\( [CrF_6]^{3-} \):
- \( F^- \) is weak field ligand → \( sp^3d^2 \) (outer orbital)
- Count = 1

\( [CoF_6]^{3-} \):
- \( F^- \) is weak field ligand → \( sp^3d^2 \)
- Count = 1

\( [Mn(CN)_6]^{3-} \):
- \( CN^- \) is strong field ligand → \( d^2sp^3 \)
- Not \( sp^3d^2 \)

\( [MnCl_6]^{3-} \):
- \( Cl^- \) is weak field ligand → \( sp^3d^2 \)
- Count = 1


Total count = 4 (\( SF_6 \), \( [CrF_6]^{3-} \), \( [CoF_6]^{3-} \), \( [MnCl_6]^{3-} \)) Quick Tip: - Strong field ligands (\( NH_3 \), \( CN^- \)) cause inner orbital hybridization (\( d^2sp^3 \)) - Weak field ligands (\( F^- \), \( Cl^- \)) promote outer orbital hybridization (\( sp^3d^2 \)) - Non-metal compounds like \( SF_6 \) always use \( sp^3d^2 \)

In intramolecular aldol condensation, two molecules of the same compound react to form a cyclic product. The compound given here contains a \(\alpha\)-hydrogen adjacent to a carbonyl group, making it a suitable candidate for the aldol reaction. Upon condensation, the compound undergoes an intramolecular aldol reaction, leading to the formation of a six-membered cyclic \(\beta\)-lactone. Option 3 is the correct structure, where the reaction forms a cyclic ester (lactone) as the major product. Quick Tip: Aldol condensations in the intramolecular form lead to the formation of cyclic products, often resulting in rings such as lactones or lactams depending on the structure of the reactants. Pay attention to the position of \(\alpha\)-hydrogens and the size of the cyclic product.

At pH 2, the amino group (-NH\(_2\)) is protonated to form an ammonium ion (-NH\(_3^+\), and the carboxyl group (-COOH) remains in its protonated form, as carboxyl acids are not deprotonated at this pH. So, compound A will have the structure: \[ H_3N - CH - COOH \]
At pH 10, the amino group remains in its deprotonated form (-NH\(_2\)) and the carboxyl group is deprotonated to form a carboxylate ion (-COO\(^-\)). Therefore, compound B will have the structure: \[ H_2N - CH - COO^- \]
Thus, the correct option is option (2). Quick Tip: In aqueous solutions, amino acids behave differently at various pH values. At low pH, the amino group is protonated and the carboxyl group remains neutral, while at high pH, the amino group is deprotonated and the carboxyl group becomes negatively charged.

The given transformation involves the conversion of ethyl benzene (C\(_6\)H\(_5\)C\(_2\)H\(_5\)) to the desired product where a Br atom is added to the benzenoid ring.

- First, bromination of ethylbenzene occurs using \( Br_2 \) in the presence of iron (Fe), which acts as a catalyst. The electrophilic aromatic substitution reaction results in the formation of a brominated product.

- After bromination, the next step involves chlorination using \( Cl_2 \) under heat (denoted by \( \Delta \)), leading to the introduction of the chlorine atom at the desired position.

- Finally, the product undergoes dehydrohalogenation in the presence of alcoholic potassium hydroxide (alc. KOH), leading to the formation of the double bond as required.

Thus, the correct set of reagents to form the desired product is option (2). Quick Tip: In aromatic substitution reactions, the presence of a halogen like chlorine or bromine can activate the ring towards further electrophilic substitution reactions. Use specific reagents and conditions (e.g., heat and alcoholic KOH) to control the outcome.

Key Concept:
Minimum boiling azeotropes form when:

Components show positive deviation from Raoult's law
Molecular interactions between unlike molecules are weaker than between like molecules
Typically occurs between molecules with different polarity or hydrogen bonding capacity


Analysis of Options:

Option 1: \( CS_2 + CH_3COCH_3 \)
- Carbon disulfide (non-polar) + acetone (polar)
- Forms minimum boiling azeotrope (shows positive deviation)

Option 2: \( H_2O + CH_3COC_2H_5 \)
- Water (strong H-bonding) + methyl ethyl ketone (weak H-bonding)
- Forms minimum boiling azeotrope

Option 3: \( C_6H_5OH + C_6H_5NH_2 \)
- Phenol + aniline (both can form strong intermolecular H-bonds)
- Shows negative deviation (forms maximum boiling azeotrope)
- Correct answer as it doesn't form minimum boiling azeotrope

Option 4: \( CH_3OH + CHCl_3 \)
- Methanol + chloroform (forms H-bonded complex)
- Shows positive deviation (minimum boiling azeotrope) Quick Tip: - Minimum boiling azeotropes: Positive deviation (weaker interactions) - Maximum boiling azeotropes: Negative deviation (stronger interactions) - Look for H-bonding capability differences between components - Similar molecules (like phenol + aniline) tend to form maximum boiling azeotropes

Concept:
The first ionization enthalpy decreases:

Down a group (atomic size increases)
From right to left across a period (effective nuclear charge decreases)


Analysis of Options:

Option 1 (87): Francium (Fr)
- Group 1, Period 7 element
- Largest atomic size in periodic table
- Lowest effective nuclear charge on valence electron
- Lowest ionization energy among given options

Option 2 (19): Potassium (K)
- Group 1, Period 4
- Higher ionization energy than Fr (smaller size)

Option 3 (32): Germanium (Ge)
- Group 14, Period 4
- Much higher ionization energy than alkali metals

Option 4 (35): Bromine (Br)
- Group 17, Period 4
- Highest ionization energy among options (high effective nuclear charge)


Periodic Trend: \[ Ionization Energy Order: Fr (87) \(<\) K (19) \(<\) Ge (32) \(<\) Br (35) \] Quick Tip: - Ionization energy \(\propto \frac{1}{atomic size}\) - Alkali metals have lowest ionization energies in their periods - Within Group 1, ionization energy decreases down the group - Francium has the lowest known ionization energy (380 kJ/mol)

Step 1: Write the balanced chemical equation \[ NaI + AgNO_3 \rightarrow AgI \downarrow + NaNO_3 \]

Step 2: Calculate moles of AgI precipitated
Molar mass of AgI = \( 108 + 127 = 235 \) g mol\(^{-1}\) \[ Moles of AgI = \frac{4.74 g}{235 g mol^{-1}} = 0.0202 mol \]

Step 3: Determine moles of NaI
From stoichiometry, 1 mole NaI produces 1 mole AgI \[ Moles of NaI = 0.0202 mol \]

Step 4: Calculate molarity of NaI solution \[ Molarity = \frac{Moles}{Volume in L} = \frac{0.0202 mol}{0.020 L} = 1.01 M \]

Rounding to nearest integer: \[ Molarity \approx 1 M \] Quick Tip: - In precipitation reactions, stoichiometry is 1:1 for simple salts - Always convert mass to moles using molar mass - Remember to convert mL to L for molarity calculations - Nearest integer rounding for final answer

Step 1: Write the equilibrium expression
For the reaction: \[ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \]
The equilibrium constant \( K_p \) is given by: \[ K_p = \frac{P_{H_2} \cdot P_{O_2}^{1/2}}{P_{H_2O}} \]

Step 2: Express partial pressures in terms of \( \alpha \)
Let initial moles of \( H_2O = 1 \)
At equilibrium:

\begin{align*
Moles of H_2O &= 1 - \alpha

\text{Moles of H_2 &= \alpha

\text{Moles of O_2 &= \alpha/2

\text{Total moles &= 1 + \alpha/2 \approx 1 \quad (\text{since \alpha \ll 1)
\end{align*

Partial pressures (total pressure = 1 bar):
\begin{align*
P_{H_2O &= (1 - \alpha) \approx 1

P_{H_2 &= \alpha

P_{O_2 &= \alpha/2
\end{align*

Step 3: Substitute into \( K_p \) expression \[ 8.0 \times 10^{-3 = \frac{\alpha \cdot (\alpha/2)^{1/2}}{1} \] \[ 8.0 \times 10^{-3} = \alpha^{3/2} / \sqrt{2} \]

Step 4: Solve for \( \alpha \) \[ \alpha^{3/2} = 8.0 \times 10^{-3} \times \sqrt{2} = 1.131 \times 10^{-2} \] \[ \alpha = (1.131 \times 10^{-2})^{2/3} = 0.049 \approx 0.05 \]

Expressed as \( \times 10^{-2} \): \[ \alpha = 5 \times 10^{-2} \] Quick Tip: - For dissociation problems, express equilibrium composition in terms of \( \alpha \) - When \( \alpha \ll 1 \), approximations simplify calculations - Remember to account for stoichiometric coefficients in \( K_p \) expression - \( K_p \) has pressure units that must balance the reaction equation

Step 1: Determine the expected enthalpy of formation using bond energies.

- Break \( X = X \) bond: \( +940 \, kJ mol^{-1} \).

- Break \( \frac{1}{2} Y = Y \) bond: \( +\frac{1}{2} \times 500 = 250 \, kJ mol^{-1} \).

- Total energy input = \( 940 + 250 = 1190 \, kJ mol^{-1} \).


Step 2: Energy released when forming bonds in \( X_2Y \).

- Assume one \( X - X \) bond (\( 410 \, kJ mol^{-1} \)) and one \( X - Y \) bond (\( 602 \, kJ mol^{-1} \)).

- Total energy released = \( 410 + 602 = 1012 \, kJ mol^{-1} \).


Step 3: Calculate expected enthalpy change.

- Expected \( \Delta H = 1190 - 1012 = 178 \, kJ mol^{-1} \).


Step 4: Calculate resonance energy.

- Given enthalpy of formation = \( 80 \, kJ mol^{-1} \).

- Resonance energy = Expected \( \Delta H \) - Actual \( \Delta H \).

- Resonance energy = \( 178 - 80 = 98 \, kJ mol^{-1} \).


Step 5: Verify.
- The resonance structures suggest partial triple bond character,
stabilizing the molecule, aligning with the calculated value.


The magnitude of the resonance energy, to the nearest integer, is \( 98 \, kJ mol^{-1} \). Quick Tip: To calculate resonance energy, subtract the enthalpy of formation from the bond dissociation enthalpy. Be sure to account for the stoichiometric coefficients when combining bond energies.

Step 1: Recall energy formula for hydrogen-like atoms
The energy of an electron in the \(n^{th}\) orbit is given by: \[ E_n = -13.6 \frac{Z^2}{n^2} eV \]
where \(Z\) = atomic number, \(n\) = principal quantum number

Step 2: Identify parameters for Be\(^{3+}\)
For Be\(^{3+}\): \[ Z = 4 \quad (Beryllium) \]
First excited state corresponds to \(n = 2\)

Step 3: Calculate energy \[ E_2 = -13.6 \frac{4^2}{2^2} = -13.6 \times \frac{16}{4} = -13.6 \times 4 = -54.4 eV \]

Step 4: Find magnitude \[ |E_2| = 54.4 eV \approx 54 eV (nearest integer) \] Quick Tip: - For hydrogen-like ions: \(E_n \propto Z^2/n^2\) - First excited state means \(n=2\) - Be\(^{3+}\) is isoelectronic with H but has \(Z=4\) - Energy becomes more negative (more stable) as \(Z\) increases

Step 1: Write Nernst equation \[ E = E^\circ - \frac{0.059}{n} \log Q \]
For the given reaction with \( n = 6 \): \[ E = 1.33 - \frac{0.059}{6} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} \right) \]

Step 2: Set E = 0 and substitute given ratio \[ 0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[H^+]^{14}} \right) \]

Step 3: Simplify the equation \[ 1.33 = \frac{0.059}{6} \left( -6 + 14 pH \right) \] \[ 1.33 = 0.059 \left( -1 + \frac{14}{6} pH \right) \] \[ 1.33 = -0.059 + 0.1377 pH \] \[ 1.389 = 0.1377 pH \] \[ pH = \frac{1.389}{0.1377} \approx 10.09 \approx 10 (nearest integer) \] Quick Tip: - For half-cell reactions, use Nernst equation: \( E = E^\circ - \frac{0.059}{n} \log Q \) - Remember \( pH = -\log[H^+] \) - When E = 0, the system is at equilibrium - Watch stoichiometric coefficients in the reaction quotient \( Q \)